[Victor E Saouma] Machanics and Design of Reinforc(Bookos.org)

7/29/2019 [Victor E Saouma] Machanics and Design of Reinforc(Bookos.org)

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DraftDRAFT

Lecture Notes in:

Mechanics and Design of

REINFORCED CONCRETE

CVEN4555

cVICTOR E. SAOUMA,

Fall 2002

Dept. of Civil Environmental and Architectural Engineering

University of Colorado, Boulder, CO 80309-0428

December 26, 2002


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Victor Saouma Mechanics and Design of Reinforced Concrete


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Contents

1 INTRODUCTION 111.1 Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.1 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.1.1.1 Mix Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.1.1.1.1 Constituents . . . . . . . . . . . . . . . . . . . . . . . . 111.1.1.1.2 Preliminary Considerations . . . . . . . . . . . . . . . . 15

1.1.1.1.3 Mix procedure . . . . . . . . . . . . . . . . . . . . . . . 151.1.1.1.4 Mix Design Example . . . . . . . . . . . . . . . . . . . 18

1.1.1.2 Mechanical Properties . . . . . . . . . . . . . . . . . . . . . . . . 191.1.2 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

1.2 Design Philosophy, USD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1141.3 Analysis vs Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1151.4 Basic Relations and Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1161.5 ACI Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

2 FLEXURE 212.1 Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

E 2-1 Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2 Section Cracked, Stresses Elastic . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2.1 Basic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2.2 Working Stress Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24E 2-2 Cracked Elastic Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25E 2-3 Working Stress Design Method; Analysis . . . . . . . . . . . . . . . . . . . 26E 2-4 Working Stress Design Method; Design . . . . . . . . . . . . . . . . . . . 27

2.3 Cracked Section, Ultimate Strength Design Method . . . . . . . . . . . . . . . . . 282.3.1 Whitney Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.3.2 Balanced Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2102.3.3 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2112.3.4 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211

2.4 Practical Design Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2122.4.1 Minimum Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2122.4.2 Beam Sizes, Bar Spacing, Concrete Cover . . . . . . . . . . . . . . . . . . 2132.4.3 Design Aids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

2.5 USD Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215E 2-5 Ultimate Strength; Review . . . . . . . . . . . . . . . . . . . . . . . . . . 215E 2-6 Ultimate Strength; Design I . . . . . . . . . . . . . . . . . . . . . . . . . . 216E 2-7 Ultimate Strength; Design II . . . . . . . . . . . . . . . . . . . . . . . . . 217


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Draft02 CONTENTSE 2-8 Exact Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

2.6 T Beams, (ACI 8.10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2202.6.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2212.6.2 Design, (balanced) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222E 2-9 T Beam; Moment Capacity I . . . . . . . . . . . . . . . . . . . . . . . . . 222

E 2-10 T Beam; Moment Capacity II . . . . . . . . . . . . . . . . . . . . . . . . . 223E 2-11 T Beam; Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

2.7 Doubly Reinforced Rectangular Beams . . . . . . . . . . . . . . . . . . . . . . . . 2262.7.1 Tests for fs and f

s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2272.7.2 Moment Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229E 2-12 Doubly Reinforced Concrete beam; Review . . . . . . . . . . . . . . . . . 230E 2-13 Doubly Reinforced Concrete beam; Design . . . . . . . . . . . . . . . . . . 232

2.8 Moment-Curvature Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2332.9 Bond & Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

2.9.1 Moment Capacity Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 239

3 SHEAR 313.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.2 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 323.3 Shear Strength of Cracked Sections . . . . . . . . . . . . . . . . . . . . . . . . . . 353.4 ACI Code Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

E 3-1 Shear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.6 Shear Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

E 3-2 Shear Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3113.7 Brackets and Corbels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3123.8 Deep Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312

4 CONTINUOUS BEAMS 414.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.2.1 Detailed Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.2.2 ACI Approximate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.3 Effective Span Design Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.4 Moment Redistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4.4.1 Elastic-Perfectly Plastic Section . . . . . . . . . . . . . . . . . . . . . . . . 444.4.2 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46E 4-1 Moment Redistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.5 Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5 ONE WAY SLABS 515.1 Types of Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2 One Way Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545.3 Design of a One Way Continuous Slab . . . . . . . . . . . . . . . . . . . . . . . . 55



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6.1 Control of Cracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61E 6-1 Crack Width . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

6.2 Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 636.2.1 Short Term Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

6.2.2 Long Term Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65E 6-2 Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

7 APPROXIMATE FRAME ANALYSIS 717.1 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 717.2 Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

7.2.1 Portal Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74E 7-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal

Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

8 COLUMNS 81

9 COLUMNS 919.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

9.1.1 Types of Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 919.1.2 Possible Arrangement of Bars . . . . . . . . . . . . . . . . . . . . . . . . . 92

9.2 Short Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 929.2.1 Concentric Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 929.2.2 Eccentric Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

9.2.2.1 Balanced Condition . . . . . . . . . . . . . . . . . . . . . . . . . 939.2.2.2 Tension Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 959.2.2.3 Compression Failure . . . . . . . . . . . . . . . . . . . . . . . . . 96

9.2.3 ACI Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

9.2.4 Interaction Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 979.2.5 Design Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97E 9-1 R/C Column, c known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97E 9-2 R/C Column, e known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99E 9-3 R/C Column, Using Design Charts . . . . . . . . . . . . . . . . . . . . . . 9139.2.6 Biaxial Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914E 9-4 Biaxially Loaded Column . . . . . . . . . . . . . . . . . . . . . . . . . . . 917

9.3 Long Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9189.3.1 Euler Elastic Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9189.3.2 Effective Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9199.3.3 Moment Magnification Factor; ACI Provisions . . . . . . . . . . . . . . . 921E 9-5 Long R/C Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 924E 9-6 Design of Slender Column . . . . . . . . . . . . . . . . . . . . . . . . . . . 925

10 PRESTRESSED CONCRETE 10110.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

10.1.1 Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10110.1.2 Prestressing Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10410.1.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10410.1.4 Tendon Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104



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Draft04 CONTENTS10.1.5 Equivalent Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10410.1.6 Load Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

10.2 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106E 10-1 Prestressed Concrete I Beam . . . . . . . . . . . . . . . . . . . . . . . . . 108

10.3 Case Study: Walnut Lane Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . 1010

10.3.1 Cross-Section Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101210.3.2 Prestressing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101210.3.3 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101310.3.4 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1013



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List of Figures

1.1 Schematic Representation of Aggregate Gradation . . . . . . . . . . . . . . . . . 121.2 MicroCracks in Concrete under Compression . . . . . . . . . . . . . . . . . . . . 1101.3 Concrete Stress Strain Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1111.4 Modulus of Rupture Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1111.5 Split Cylinder (Brazilian) Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1111.6 Biaxial Strength of Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1121.7 Time Dependent Strains in Concrete . . . . . . . . . . . . . . . . . . . . . . . . . 113

2.1 Strain Diagram Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2 Transformed Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 Stress Diagram Cracked Elastic Section . . . . . . . . . . . . . . . . . . . . . . . 232.4 Desired Stress Distribution; WSD Method . . . . . . . . . . . . . . . . . . . . . . 242.5 Cracked Section, Limit State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.6 Whitney Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2102.7 Bar Spacing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2152.8 T Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2202.9 T Beam as Rectangular Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2202.10 T Beam Strain and Stress Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 221

2.11 Decomposition of Steel Reinforcement for T Beams . . . . . . . . . . . . . . . . . 2212.12 Doubly Reinforced Beams; Strain and Stress Diagrams . . . . . . . . . . . . . . . 2262.13 Different Possibilities for Doubly Reinforced Concrete Beams . . . . . . . . . . . 2272.14 Strain Diagram, Doubly Reinforced Beam; is As Yielding? . . . . . . . . . . . . . 2272.15 Strain Diagram, Doubly Reinforced Beam; is As Yielding? . . . . . . . . . . . . . 2282.16 Summary of Conditions for top and Bottom Steel Yielding . . . . . . . . . . . . . 2292.17 Be n d in g of a Be am . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2342.18 Moment-Curvature Relation for a Beam . . . . . . . . . . . . . . . . . . . . . . . 2352.19 Bond and Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2362.20 Actual Bond Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2372.21 Splitting Along Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2372.22 Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2382.23 Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2382.24 Hooks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2402.25 Bar cutoff requirements of the ACI code . . . . . . . . . . . . . . . . . . . . . . . 2412.26 Standard cutoff or bend points for bars in approximately equal spans with uni-

formly distributed load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2422.27 Moment Capacity Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

3.1 Principal Stresses in Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31


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Draft02 LIST OF FIGURES3.2 Types of Shear Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.3 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 323.4 Mohrs Circle for Shear Strength of Uncracked Section . . . . . . . . . . . . . . . 333.5 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 343.6 Free Body Diagram of a R/C Section with a Flexural Shear Crack . . . . . . . . 35

3.7 Equilibrium of Shear Forces in Cracked Section . . . . . . . . . . . . . . . . . . . 363.8 Summary of ACI Code Requirements for Shear . . . . . . . . . . . . . . . . . . . 373.9 Corbel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.10 Shear Friction Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3103.11 Shear Friction Across Inclined Reinforcement . . . . . . . . . . . . . . . . . . . . 310

4.1 Continuous R/C Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2 Load Positioning on Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . 414.3 ACI Approximate Moment Coefficients . . . . . . . . . . . . . . . . . . . . . . . . 434.4 Design Negative Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.5 Moment Diagram of a Rigidly Connected Uniformly Loaded Beam . . . . . . . . 454.6 Moment Curvature of an Elastic-Plastic Section . . . . . . . . . . . . . . . . . . . 45

4.7 Plastic Moments in Uniformly Loaded Rigidly Connected Beam . . . . . . . . . . 454.8 Plastic Redistribution in Concrete Sections . . . . . . . . . . . . . . . . . . . . . 464.9 Block Diagram for R/C Building Design . . . . . . . . . . . . . . . . . . . . . . . 48

5.1 Types of Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2 One vs Two way slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.3 Load Distribution in Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.4 Load Transfer in R/C Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6.1 Crack Width Equation Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 626.2 Uncracked Transformed and Cracked Transformed X Sections . . . . . . . . . . . 646.3 Time Dependent Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 656.4 Time Dependent Strain Distribution . . . . . . . . . . . . . . . . . . . . . . . . . 666.5 Short and long Term Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

7.1 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments . 727.2 Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces737.3 Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments 737.4 Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear . . . 757.5 Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment . . 757.6 Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force767.7 Example; Approximate Analysis of a Building . . . . . . . . . . . . . . . . . . . . 777.8 Approximate Analysis of a Building; Moments Due to Vertical Loads . . . . . . . 79

7.9 Approximate Analysis of a Building; Shears Due to Vertical Loads . . . . . . . . 7107.10 Approximate Analysis for Vertical Loads; Spread-Sheet Format . . . . . . . . . . 7127.11 Approximate Analysis for Vertical Loads; Equations in Spread-Sheet . . . . . . . 7137.12 Approximate Analysis of a Building; Moments Due to Lateral Loads . . . . . . . 7147.13 Portal Method; Spread-Sheet Format . . . . . . . . . . . . . . . . . . . . . . . . . 7167.14 Portal Method; Equations in Spread-Sheet . . . . . . . . . . . . . . . . . . . . . . 717

9.1 Types of columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 919.2 Tied vs Spiral Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91



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DraftLIST OF FIGURES 039.3 Possible Bar arrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 929.4 Sources of Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 939.5 Load Moment Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 939.6 Strain and Stress Diagram of a R/C Column . . . . . . . . . . . . . . . . . . . . 949.7 Column Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

9.8 Failure Surface of a Biaxially Loaded Column . . . . . . . . . . . . . . . . . . . . 9149.9 Load Contour at Plane of Constant Pn, and Nondimensionalized Corresponding

plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9159.10 Biaxial Bending Interaction Relations in terms of . . . . . . . . . . . . . . . . . 9169.11 Bilinear Approximation for Load Contour Design of Biaxially Loaded Columns . 9169.12 Euler Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9189.13 Column Failures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9199.14 Critical lengths of columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9209.15 Effective length Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9219.16 Standard Alignment Chart (ACI) . . . . . . . . . . . . . . . . . . . . . . . . . . . 9229.17 Minimum Column Eccentricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9229.18 P-M Magnification Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . 923

10.1 Pretensioned Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . 10210.2 Posttensioned Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . 10210.3 7 Wire Prestressing Tendon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10310.4 Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978)10510.5 Determination of Equivalent Loads . . . . . . . . . . . . . . . . . . . . . . . . . . 10510.6 Load-Deflection Curve and Corresponding Internal Flexural Stresses for a Typi-

cal Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . . . . . . . 10610.7 Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum

Moment Section and Support Section, (Nilson 1978) . . . . . . . . . . . . . . . . 10710.8 Walnut Lane Bridge, Plan View . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011

10.9 Walnut Lane Bridge, Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . 1012



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Draft04 LIST OF FIGURES



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Draft

List of Tables

1.1 ASTM Sieve Designations Nominal Sizes Used for Concrete Aggregates . . . . . 131.2 ASTM C33 Grading Limits for Coarse Concrete Aggregates . . . . . . . . . . . . 131.3 ASTM C33 Grading Limits for Fine Concrete Aggregates . . . . . . . . . . . . . 131.4 Example of Fineness Modulus Determination for Fine Aggregate . . . . . . . . . 151.5 Recommended Slumps (inches) for Various Types of Construction . . . . . . . . 161.6 Recommended Average Total Air Content as % of Different Nominal Maximum

Sizes of Aggregates and Levels of Exposure . . . . . . . . . . . . . . . . . . . . . 161.7 Approximate Mixing Water Requirements, lb/yd3 of Concrete For Different

Slumps and Nominal Maximum Sizes of Aggregates . . . . . . . . . . . . . . . . . 171.8 Relationship Between Water/Cement Ratio and Compressive Strength . . . . . . 171.9 Volume of Dry-Rodded Coarse Aggregate per Unit Volume of Concrete for Dif-

ferent Fineness Moduli of Sand . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.10 Creep Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1131.11 Properties of Reinforcing Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1141.12 Strength Reduction Factors, . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

2.1 Total areas for various numbers of reinforcing bars (inch2) . . . . . . . . . . . . . 2142.2 Minimum Width (inches) according to ACI Code . . . . . . . . . . . . . . . . . . 214

4.1 Building Structural Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5.1 Recommended Minimum Slab and Beam Depths . . . . . . . . . . . . . . . . . . 54

7.1 Columns Combined Approximate Vertical and Horizontal Loads . . . . . . . . . 7187.2 Girders Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . 719


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Draft

Chapter 1

INTRODUCTION

1.1 Material

1.1.1 Concrete

This section is adapted from Concrete by Mindess and Young, Prentice Hall, 1981

1.1.1.1 Mix Design

1.1.1.1.1 Constituents

1 Concrete is a mixture of Portland cement, water, and aggregates (usually sand and crushedstone).

2 Portland cement is a mixture of calcareous and argillaceous materials which are calcined ina kiln and then pulverized. When mixed with water, cement hardens through a process called

hydration.3 Ideal mixture is one in which:

1. A minimum amount of cement-water paste is used to fill the interstices between theparticles of aggregates.

2. A minimum amount of water is provided to complete the chemical reaction with cement.Strictly speaking, a water/cement ratio of about 0.25 is needed to complete this reaction,but then the concrete will have a very low workability.

In such a mixture, about 3/4 of the volume is constituted by the aggregates, and the remaining1/4 being the cement paste.

4 Smaller particles up to 1/4 in. in size are called fine aggregates, and the larger ones beingcoarse aggregates.

5 Portland Cement has the following ASTM designation

I Normal

II Moderate sulfate resistant, moderate heat of hydration

III High early strength (but releases too much heat)


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Draft1.1 Material 13ASTM Size

Design. mm in.

Coarse Aggregate

3 in. 75 3

21/2 in. 63 2.5

2 in. 50 211/2 in. 37.5 1.5

1 in. 25 13/4 in. 19 0.751/2 in. 12.5 0.503/8 in. 9.5 0.375

Fine Aggregate

No. 4 4.75 0.187No. 8 2.36 0.0937

No. 16 1.18 0.0469No. 30 0.60 (600 m) 0.0234

No. 50 300 m 0.0124No. 100 150 m 0.0059

Table 1.1: ASTM Sieve Designations Nominal Sizes Used for Concrete Aggregates

Sieve Size % Passing Each Sieve(Nominal Maximum Size)

11/2 in. 1 in. 3/4 in. 1/2 in.

11/2 in. 95-100 100 - -1 in. - 95-100 100 -

3/4 in. 35-70 - 90-100 1001/2 in. - 25-60 - 90-1003/8 in. 10-30 - 20-55 40-70No. 4 0-5 0-10 0-10 0-15No. 8 - 0-5 0-5 0-5

Table 1.2: ASTM C33 Grading Limits for Coarse Concrete Aggregates

Sieve Size % Passing

3/4 in. 100

No. 4 95-100No. 8 80-100No. 16 50-85No. 30 25-60No. 50 10-30No. 100 2-10

Table 1.3: ASTM C33 Grading Limits for Fine Concrete Aggregates



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Draft14 INTRODUCTION16 Moisture states are defined as

Oven-dry (OD): all moisture is removed from the aggregate.

Air-dry (AD): all moisture is removed from the surface, but internal pores are partially full.

Saturated-surface-dry (SSD): All pores are filled with water, but no film of water on thesurface.

Wet: All pores are completely filled with a film of water on the surface.

17 Based on the above, we can determine

Absorption capacity (AC): is the maximum amount of water the aggregate can absorb

AC =WSSD WOD

WOD 100% (1.1)

most normal -weight aggregates (fine and coarse) have an absorption capacity in the rangeof 1% to 2%.

Surface Moisture (SM): is the water in excess of the SSD state

SM =WWet WSSD

WSSD 100% (1.2)

18 The fineness modulus is a parameter which describe the grading curve and it can be usedto check the uniformity of the grading. It is usually computed for fine aggregates on the basisof

F.M. = cumulative percent retained on standard sieves

100

(1.3)

where the standard sieves used are No. 100, No. 50, No. 30, No. 16, No. 8, and No. 4, and3/8 in, 3/4 in, 11/2 in and larger.

19 The fineness modulus for fine aggregate should lie between 2.3 and 3.1 A small numberindicates a fine grading, whereas a large number indicates a coarse material.

20 Table 1.4 illustrates the determination of the fineness modulus.

21 Fineness modulus of fine aggregate is required for mix proportioning since sand gradationhas the largest effect on workability. A fine sand (low fineness modulus) has much higher pasterequirements for good workability.

22

The fineness modulus of coarse aggregate is not used for mix design purposes.23 no-fines concrete has little cohesiveness in the fresh state and can not be compacted to avoid-free condition. Hence, it will have a low strength, high permeability. Its only advantage islow density, and high thermal insulation which can be used if structural requirements are nothigh.



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Draft1.1 Material 15Sieve Weight Amount Cumulative CumulativeSize Retained Retained Amount Amount

(g) (wt. %) Retained (%) Passing (%)

No. 4 9 2 2 98No. 8 46 9 11 89

No. 16 97 19 30 70No. 30 99 20 50 50No. 50 120 24 74 26No. 100 91 18 92 8

Sample Weight 500 g.

= 259

Fineness modulus=259/100=2.59

Table 1.4: Example of Fineness Modulus Determination for Fine Aggregate

1.1.1.1.2 Preliminary Considerations

24 There are two fundamental aspects to mix design to keep in mind:

1. Water/Cement ratio: where the strength is inversely proportional to the water to cementratio, approximately expressed as:

fc =A

B1.5w/c(1.4)

For fc in psi, A is usually taken as 14,000 and B depends on the type of cement, but maybe taken to be about 4. It should be noted that w/c controls not only the strength, butalso the porosity and hence the durability.

2. Aggregate Grading: In order to minimize the amount of cement paste, we must maximizethe volume of aggregates. This can be achieved through proper packing of the granularmaterial. The ideal grading curve (with minimum voids) is closely approximated bythe Fuller curve

Pt =

d

D

q(1.5)

where Pt is the fraction of total solids finer than size d, and D is the maximum particlesize, q is generally taken as 1/2, hence the parabolic grading.

1.1.1.1.3 Mix procedure

25 Before starting the mix design process, the following material properties should be deter-mined:

1. Sieve analysis of both fine and coarse aggregates

2. Unit weight of the coarse aggregate

3. Bulk specific gravities



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Draft16 INTRODUCTION4. absorption capacities of the aggregates

1. Slump1 must be selected for the particular job to account for the anticipated methodof handling and placing concrete, Table 1.5 As a general rule, adopt the lowest possible

Type of Construction Max MinFoundation walls and footings 3 1Plain footings, caissons 3 1Beams and reinforced walls 4 1Building columns 4 1Pavement and slabs 3 1Mass concrete 3 1

Table 1.5: Recommended Slumps (inches) for Various Types of Construction

slump.

2. Maximum aggregate size: in general the largest possible size should be adopted.However, it should be noted that:

(a) For reinforced concrete, the maximum size may not exceed one-fifth of the mini-mum dimensions between the forms, or three-fourths of the minimum clear spacingbetween bars, or between steel and forms.

(b) For slabs on grade, the maximum size may not exceed one-third the slab depth.

In general maximum aggregate size is 3/4 in or 1 in.

3. Water and Air content Air content will affect workability (some time it is better to

increase air content rather than increasing w/c which will decrease strength). Air contentcan be increased through the addition of admixtures. Table 1.6 tabulates recommendedvalues of air content (obtained through such admixtures) for different conditions (forinstance under severe freezing/thawing air content should be high).

Recommended water requirements are given by Table 1.7.

Sizes of Aggregates

Exposure 3/8 in. 1/2 in. 3/4 in. 1 in. 11/2 in.

Mild 4.5 4.0 3.5 3.5 3.0Moderate 6.0 5.5 5.0 4.5 4.4Extreme 7.5 7.0 6.0 6.05 5.5

Table 1.6: Recommended Average Total Air Content as % of Different Nominal Maximum Sizesof Aggregates and Levels of Exposure

1The slump test (ASTM C143) is a measure of the shear resistance of concrete to flowing under its own weight.It is a good indicator of the concrete workability. A hollow mold in the form of a frustum of a cone is filledwith concrete in three layers of equal volume. Each layer is rodded 25 times. The mold is then lifted vertically,and the slump is measured by determining the difference between the height of the mold and the height of theconcrete over the original center of the base of the specimen.



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Draft1.1 Material 17Slump Sizes of Aggregates

in. 3/8 in. 1/2 in. 3/4 in. 1 in. 11/2 in.

Non-Air-Entrained Concrete

1-2 350 335 315 300 2753-4 385 365 340 325 300

6-7 410 385 360 340 315Air-Entrained Concrete

1-2 305 295 280 270 2503-4 340 325 305 295 2756-7 365 345 325 310 290

Table 1.7: Approximate Mixing Water Requirements, lb/yd3 of Concrete For Different Slumpsand Nominal Maximum Sizes of Aggregates

4. Water/cement ratio: this is governed by both strength and durability. Table 1.8

provides some guidance in terms of strength.

28 days w/c Ratio by Weightfc Non-air-entrained Air-entrained

6,000 0.41 -5,000 0.48 0.404,000 0.57 0.483,000 0.68 0.592,000 0.82 0.74

Table 1.8: Relationship Between Water/Cement Ratio and Compressive Strength

For durability, if there is a severe exposure (freeze/thaw, exposure to sea-water, sulfates),then there are severe restrictions on the W/C ratio (usually to be kept just under 0.5)

5. Cement Content: Once the water content and the w/c ratio are determined, the amountof cement per unit volume of concrete is determined simply by dividing the estimatedwater requirement by the w/c ratio.

6. Coarse Aggregate Content: Volume of coarse aggregate required per cubic yard ofconcrete depends on its maximum size and the fineness modulus of the fine aggregate,Table 1.9. The oven dry (OD) volume of coarse aggregate in ft3 required per cubic yard

is simply equal to the value from Table 1.9 multiplied by 27. This volume can then beconverted to an OD weight by multiplying it by the dry-rodded2 weight per cubic foot ofcoarse aggregate.

7. The fine aggregate content can be estimated by subtracting the volume of cement,water, air and coarse aggregate from the total volume. The weight of the fine aggregatecan then be obtained by multiplying this volume by the density of the fine aggregate.

2Dry Rodded Volume (DRV) is the normal volume of space a material occupies.



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Draft18 INTRODUCTIONAgg. Size Sand Fineness Moduli

in 2.40 2.60 2.80 3.00

3/8 0.50 0.48 0.46 0.441/2 0.59 0.57 0.55 0.533/4 0.66 0.64 0.62 0.60

1 0.71 0.69 0.67 0.6511/2 0.76 0.74 0.72 0.70

Table 1.9: Volume of Dry-Rodded Coarse Aggregate per Unit Volume of Concrete for DifferentFineness Moduli of Sand

8. Adjustment for moisture in the aggregates: is necessary. If aggregates are airdry, they will absorb some water (thus effectively lowering the w/c), or if aggregates aretoo wet they will release water (increasing the w/c and the workability but reducing thestrength).

1.1.1.1.4 Mix Design Example

Concrete is required for an exterior column to be located above ground in an area wheresubstantial freezing and thawing may occur. The concrete is required to have an average 28-day compressive strength of 5,000 psi. For the conditions of placement, the slump should bebetween 1 and 2 in, the maximum aggregate size should not exceed 3/4 in. and the propertiesof the materials are as follows:

Cement: Type I specific gravity = 3.15

Coarse Aggregates: Bulk specific gravity (SSD) = 2.70; absorption capacity= 1.0%; Total

moisture content = 2.5%; Dry-rodded unit weight = 100 lb/ft3

Fine Aggregates: Bulk specific gravity (SSD) = 2.65; absorption capacity = 1.3 %; Totalmoisture content=5.5%; fineness modulus = 2.70

The sieve analyses of both the coarse and fine aggregates fall within the specified limits. Withthis information, the mix design can proceed:

1. Choice ofslump is consistent with Table 1.5.

2. Maximum aggregate size (3/4 in) is governed by reinforcing details.

3. Estimation of mixing water: Because water will be exposed to freeze and thaw, it must

be air-entrained. From Table 1.6 the air content recommended for extreme exposure is6.0%, and from Table 1.7 the water requirement is 280 lb/yd3

4. From Table 1.8, the water to cement ratio estimate is 0.4

5. Cement content, based on steps 4 and 5 is 280/0.4=700 lb/yd3

6. Coarse aggregate content, interpolating from Table 1.9 for the fineness modulus ofthe fine aggregate of 2.70, the volume of dry-rodded coarse aggregate per unit volume ofconcrete is 0.63. Therefore, the coarse aggregate will occupy 0.63 27 = 17.01 ft3/yd3.



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Draft1.1 Material 19The OD weight of the coarse aggregate is 17.01 ft3/yd3, 100 lbs/ft3=1,701 lb. The SSDweight is 1,701 1.01=1,718 lb.

7. Fine aggregate content Knowing the weights and specific gravities of the water, cement,and coarse aggregate, and knowing the air volume, we can calculate the volume per yd3

occupied by the different ingredients.

Water 280/62.4 = 4.49 ft3

Cement 700/(3.15)(62.4) = 3.56 ft3

Coarse Aggregate (SSD) 1,718/(2.70)(62.4) = 1.62 ft3

Air (0.06)(27) = 1.62 ft3

19.87 ft3

Hence, the fine aggregate must occupy a volume of 27.0 19.87 = 7.13 ft3. The requiredSSD weight of the fine aggregate is 7.13 ft3 (2.65)(62.4)lb/ft3 =1,179 lbs lb.

8. Adjustment for moisture in the aggregate. Since the aggregate will be neither SSD orOD in the field, it is necessary to adjust the aggregate weights for the amount of watercontained in the aggregate. Only surface water need be considered; absorbed water doesnot become part of the mix water. For the given moisture contents, the adjusted aggre-gate weights become:

Coarse aggregate (wet)=1,718(1.025-0.01) = 1,744 lb/yd3 of dry coarseFine aggregate (wet)=1,179(1.055-0.013) = 1,229 lb/yd3 of dry fine

Surface moisture contributed by the coarse aggregate is 2.5-1.0 = 1.5%; by the fine ag-gregate: 5.5-1.3 = 4.2%; Hence we need to decrease water to280-1,718(0.015)-1,179(0.042) = 205 lb/yd3.

Thus, the estimated batch weight per yd3 are

Water 205 lbCement 700 lbWet coarse aggregate 1,744 lbWet fine aggregate 1,229 lb

3,878 lb/yd33,87827 143.6 lb/ft

3

1.1.1.2 Mechanical Properties

26 Contrarily to steel to modulus of elasticity of concrete depends on the strength and is given

by

E = 57, 000

fc (1.6)

or

E = 331.5

fc (1.7)

where both fc and E are in psi and is in lbs/ft3.



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Draft110 INTRODUCTION27 Normal weight and lightweight concrete have equal to 150 and 90-120 lb/ft3 respectively.

28 Poissons ratio = 0.15.

29 Typical concrete (compressive) strengths range from 3,000 to 6,000 psi; However high strengthconcrete can go up to 14,000 psi.

30 Stress-strain curve depends on

1. Properties of aggregates

2. Properties of cement

3. Water/cement ratio

4. Strength

5. Age of concrete

6. Rate of loading, as rate

, strength

31 Non-linear part of stress-strain curve is caused by micro-cracking around the aggregates, Fig.1.2

~ 0.5 fc

u

fc

Linear

Non-Linear

Figure 1.2: MicroCracks in Concrete under Compression

32 Irrespective of fc, maximum strain under compression is 0.003, Fig. 1.333 Full strength of concrete is achieved in about 28 days

fct =t

4.0 + .85tfc,28 (1.8)

or

t (days) 1 2 4 7 10 15

%fc,28 20 35 54 70 80 90

34 Concrete always gain strength in time, but a decreasing rate



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Draft1.1 Material 111

u

=

fc

fc

0.003

/2

Figure 1.3: Concrete Stress Strain Curve

35 The tensile strength of concrete ft is very difficult to measure experimentally. Acceptedvalues

f

t 0.07 0.11f

c (1.9-a) 3 5

fc (1.9-b)

36 Rather than the tensile strength, it is common to measure the modulus of rupture fr, Fig.1.4

0 00 00 01 11 11 1 0 00 00 01 11 11 1Figure 1.4: Modulus of Rupture Test

Figure 1.5: Split Cylinder (Brazilian) Test

fr 7.5

fc (1.10)



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Draft112 INTRODUCTION

fc

fc

~ 20% increase in strength

f

f

t

t

11

2

2

1

2

Figure 1.6: Biaxial Strength of Concrete

37 Using split cylinder (or brazilian test), Fig. 1.5 ft 68

fc. For this test, a nearly uniformtensile stress

=2P

dt(1.11)

where P is the applied compressive load at failure, d and t are diameter and thickness of thespecimen respectively.

38 In most cases, concrete is subjected to uniaxial stresses, but it is possible to have biaxial(shells, shear walls) or triaxial (beam/column connections) states of stress.

39 Biaxial strength curve is shown in Fig. 1.6

40 Concrete has also some time-dependent properties

Shrinkage: when exposed to air (dry), water tends to evaporate from the concrete surface, shrinkage. It depends on the w/c and relative humidity. sh 0.00020.0007. Shrinkagecan cause cracking if the structure is restrained, and may cause large secondary stresses.

If a simply supported beam is fully restrained against longitudinal deformation, then

sh = Esh (1.12-a)

= 57, 000

3, 000(0.0002) = 624 psi >3, 000

10

ft(1.12-b)

if the concrete is restrained, then cracking will occur3.

Creep: can be viewed as the squeezing out of water due to long term stresses (analogous toconsolidation in clay), Fig. 1.7.

3For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement,can reduce the shrinkage by 75%.



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Draft1.1 Material 113

no load constant load

creepElastic recovery

Residual

Creep recovery

no load

Figure 1.7: Time Dependent Strains in Concrete

Creep coefficient, Table 1.10

Cu =ctci 2 3 (1.13-a)

Ct =t0.6

10 + t0.6Cu (1.13-b)

fc 3,000 4,000 6,000 8,000

Cu 3.1 2.9 2.4 2.0

Table 1.10: Creep Coefficients

41 Coefficient of thermal expansion is 0.65 105 /deg F for normal weight concrete.

1.1.2 Reinforcing Steel

42 Steel is used as reinforcing bars in concrete, Table 1.11.

43 Bars have a deformation on their surface to increase the bond with concrete, and usuallyhave a yield stress of 60 ksi.

44 Maximum allowable fy is 80 ksi.45 Stirrups, used as vertical reinforcement to resist shear, usually have a yield stress of only 40ksi

46 Steel loses its strength rapidly above 700 deg. F (and thus must be properly protected fromfire), and becomes brittle at 30 deg. F47 Prestressing Steel cables have an ultimate strength up to 270 ksi.



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Draft114 INTRODUCTIONBar Designation Diameter Area Perimeter Weight

(in.) ( in2) in lb/ft

No. 2 2/8=0.250 0.05 0.79 0.167No. 3 3/8=0.375 0.11 1.18 0.376No. 4 4/8=0.500 0.20 1.57 0.668

No. 5 5/8=0.625 0.31 1.96 1.043No. 6 6/8=0.750 0.44 2.36 1.5202No. 7 7/8=0.875 0.60 2.75 2.044No. 8 8/8=1.000 0.79 3.14 2.670No. 9 9/8=1.128 1.00 3.54 3.400

No. 10 10/8=1.270 1.27 3.99 4.303No. 11 11/8=1.410 1.56 4.43 5.313

No. 14 14/8=1.693 2.25 5.32 7.650

No. 18 18/8=2.257 4.00 7.09 13.60

Table 1.11: Properties of Reinforcing Bars

48 Welded wire fabric is often used to reinforce slabs and shells. It has both longitudinal andtransverse cold-drawn steel. They are designated by AAW BB, such as 66W1.41.4where spacing of the wire is 6 inch, and a cross section of 0.014 in2.

1.2 Design Philosophy, USD

49 ACI refers to this method as the Strength Design Method, (previously referred to as theUltimate Strength Method).

Rn iQi (1.14)

where

is a strength reduction factor, less than 1, and must account for the type of structuralelement, Table 1.12 (ACI 9.3.2)

Type of Member

Axial Tension 0.9Flexure 0.9Axial Compression, spiral reinforcement 0.75

Axial Compression, other 0.70Shear and Torsion 0.85Bearing on concrete 0.70

Table 1.12: Strength Reduction Factors,

Rn is the nominal resistance (or strength).



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Draft1.3 Analysis vs Design 115Ru = Rd = Rn is the design strength.

i is the load factor corresponding to Qi and is greater than 1.

iQi is the required strength based on the factored load:

i is the type of load

Mn Mu (1.15-a)Vn Vu (1.15-b)Pn Pu (1.15-c)

50 Note that the subscript d and u are equivalent.

51 The various factored load combinations which must be considered (ACI: 9.2) are

1. 1.4D+1.7L

2. 0.75(1.4D+1.7L+1.7W)3. 0.9D+1.3W

4. 1.05D+1.275W

5. 0.9D+1.7H

6. 1.4D +1.7L+1.7H

7. 0.75(1.4D+1.4T+1.7L)

8. 1.4(D+T)

where D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature;

H= soil. We must select the one with the largest limit state load.52 Serviceability Limit States must be assessed under service loads (not factored). Themost important ones being

1. Deflections

2. Crack width (for R/C)

3. Stability

1.3 Analysis vs Design

53 In R/C we always consider one of the following problems:

Analysis: Given a certain design, determine what is the maximum moment which can beapplied.

Design: Given an external moment to be resisted, determine cross sectional dimensions (b andh) as well as reinforcement (As). Note that in many cases the external dimensions of thebeam (b and h) are fixed by the architect.

54 We often consider the maximum moment along a member, and design accordingly.



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Draft116 INTRODUCTION1.4 Basic Relations and Assumptions

55 In developing a design/analysis method for reinforced concrete, the following basic relationswill be used:

1. Equilibrium: of forces and moment at the cross section. 1) Fx = 0 or Tension in thereinforcement = Compression in concrete; and 2) M = 0 or external moment (that is theone obtained from the moment envelope) equal and opposite to the internal one (tensionin steel and compression of the concrete).

2. Material Stress Strain: We recall that all normal strength concrete have a failure strainu = .003 in compression irrespective of f

c.

56 Basic assumptions used:

Compatibility of Displacements: Perfect bond between steel and concrete (no slip). Notethat those two materials do also have very close coefficients of thermal expansion under

normal temperature.Plane section remain plane strain is proportional to distance from neutral axis.Neglect tensile strength in all cases.

1.5 ACI Code

Attached is an unauthorized copy of some of the most relevant ACI-318-89 design code provi-sions.

8.1.1 - In design of reinforced concrete structures, members shall be proportioned for ad-equate strength in accordance with provisions of this code, using load factors and strength

reduction factors specified in Chapter 9.8.3.1 - All members of frames or continuous construction shall be designed for the maximum

effects of factored loads as determined by the theory of elastic analysis, except as modifiedaccording to Section 8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used.

8.5.1 - Modulus of elasticity Ec for concrete may be taken as W1.5c 33

fc ( psi) for values

of Wc between 90 and 155 lb per cu ft. For normal weight concrete, Ec may be taken as57, 000

fc.

8.5.2 - Modulus of elasticity Es for non-prestressed reinforcement may be taken as 29,000psi.

9.1.1 - Structures and structural members shall be designed to have design strengths at allsections at least equal to the required strengths calculated for the factored loads and forces in

such combinations as are stipulated in this code.9.2 - Required Strength9.2.1 - Required strength U to resist dead load D and live load L shall be at least equal to

U = 1.4D + 1.7L (1.16)

9.2.2 - If resistance to structural effects of a specified wind load W are included in design,the following combinations of D, L, and W shall be investigated to determine the greatestrequired strength U

U = 0.75(1.4D + 1.7L + 1.7W) (1.17)



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Draft1.5 ACI Code 117where load combinations shall include both full value and zero value of L to determine the moresevere condition, and

U = 0.9D + 1.3W (1.18)

but for any combination of D, L, and W, required strength U shall not be less than Eq. (9-1).9.3.1 - Design strength provided by a member, its connections to other members, and its

cross sections, in terms of flexure, axial load, shear, and torsion, shall be taken as the nominalstrength calculated in accordance with requirements and assumptions of this code, multipliedby a strength reduction factor .

9.3.2 - Strength reduction factor shall be as follows:9.3.2.1 - Flexure, without axial load 0.909.4 - Design strength for reinforcement Designs shall not be based on a yield strength of

reinforcement fy in excess of 80,000 psi, except for prestressing tendons.10.2.2 - Strain in reinforcement and concrete shall be assumed directly proportional to

the distance from the neutral axis, except, for deep flexural members with overall depth toclear span ratios greater than 2/5 for continuous spans and 4/5 for simple spans, a non-lineardistribution of strain shall be considered. See Section 10.7.

10.2.3 - Maximum usable strain at extreme concrete compression fiber shall be assumedequal to 0.003.

10.2.4 - Stress in reinforcement below specified yield strength fy for grade of reinforcementused shall be taken as Es times steel strain. For strains greater than that corresponding to fy,stress in reinforcement shall be considered independent of strain and equal to fy.

10.2.5 - Tensile strength of concrete shall be neglected in flexural calculations of reinforcedconcrete, except when meeting requirements of Section 18.4.

10.2.6 - Relationship between concrete compressive stress distribution and concrete strainmay be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results inprediction of strength in substantial agreement with results of comprehensive tests.

10.2.7 - Requirements of Section 10.2.5 may be considered satisfied by an equivalent rect-

angular concrete stress distribution defined by the following:10.2.7.1 - Concrete stress of 0.85fc shall be assumed uniformly distributed over an equiva-lent compression zone bounded by edges of the cross section and a straight line located parallelto the neutral axis at a distance (a = 1c) from the fiber of maximum compressive strain.

10.2.7.2 - Distance c from fiber of maximum strain to the neutral axis shall be measuredin a direction perpendicular to that axis.

10.2.7.3 - Factor 1 shall be taken as 0.85 for concrete strengths f

c up to and including4,000 psi. For strengths above 4,000 psi, 1 shall be reduced continuously at a rate of 0.05 foreach 1000 psi of strength in excess of 4,000 psi, but 1 shall not be taken less than 0.65.

10.3.2 - Balanced strain conditions exist at a cross section when tension reinforcementreaches the strain corresponding to its specified yield strength fy just as concrete in compressionreaches its assumed ultimate strain of 0.003.

10.3.3 - For flexural members, and for members subject to combined flexure and compres-sive axial load when the design axial load strength (Pn) is less than the smaller of (0.10f

cAg)or (Pb), the ratio of reinforcement p provided shall not exceed 0.75 of the ratio b that wouldproduce balanced strain conditions for the section under flexure without axial load. For mem-bers with compression reinforcement, the portion of b equalized by compression reinforcementneed not be reduced by the 0.75 factor.

10.3.4 - Compression reinforcement in conjunction with additional tension reinforcementmay be used to increase the strength of flexural members.



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Draft118 INTRODUCTION10.5.1 - At any section of a flexural member, except as provided in Sections 10.5.2 and

10.5.3, where positive reinforcement is required by analysis, the ratio provided shall not beless than that given by

min =200

fy(1.19)



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Draft

Chapter 2

FLEXURE

1 This is probably the longest chapter in the notes, we shall cover in great details flexuraldesign/analysis of R/C beams starting with uncracked section to failure conditions.

1. Uncracked elastic (uneconomical)

2. cracked elastic (service stage)

3. Ultimate (failure)

2.1 Uncracked Section

h d

b

A

c

s

s

Figure 2.1: Strain Diagram Uncracked Section

2 Assuming perfect bond between steel and concrete, we have s = c, Fig. 2.1

s = c fsEs

=fcEc fs = Es

Ecfc fs = nfc (2.1)

where n is the modular ratio n = EsEc

3 Tensile force in steel Ts = Asfs = Asnfc

4 Replace steel by an equivalent area of concrete, Fig. 2.2.


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Draft22 FLEXURE

S(n-1)A

2

S(n-1)A

2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Figure 2.2: Transformed Section

5 Homogeneous section & under bending

fc =M c

I fs = nfc (2.2)

6 Make sure that +max < f

t

Example 2-1: Uncracked Section

Given fc = 4,000 psi; f

t = 475 psi; fy = 60,000 psi; M = 45 ft-k = 540,000 in-lb; As = 2.35in2

Determine f+max, f

max, and fs

s

2

t

bA = 2.35 in

25"23"

10"

y

y

Solution:

n =29, 000

57

4, 000= 8 (n 1)As = (8 1)(2.35) = 16.45 in2 (2.3-a)

yb =

(10)(25)( 252 ) + (16.45)(2)

(25)(10) + 16.45 (2.3-b)

yb = 11.8 in (2.3-c)

yt = 25 11.8 = 13.2 in (2.3-d)I =

(10)(25)3

12+ (25)(10)(13.2 12.5)2 + (16.45)(23 13.2)2 (2.3-e)

= 14, 722 in2 (2.3-f)

fcc =M c

I=

(540, 000) lb.in(13.2)in

(14, 722) in4= 484 psi (2.3-g)



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Draft2.2 Section Cracked, Stresses Elastic 23fct =

M c

I=

(540, 000) lb.in(25 13.2) in(14, 722) in4

= 433 psi < 475 psi

(2.3-h)

fs = nM c

I= (8)

(540, 000)(23 13.2) in(14, 722)

= 2, 876 psi (2.3-i)

2.2 Section Cracked, Stresses Elastic

7 This is important not only as an acceptable alternative ACI design method, but also for thelater evaluation of crack width under service loads.

2.2.1 Basic Relations

8 If fct > fr, fcc


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Draft24 FLEXURE2.2.2 Working Stress Method

12 Referred to as Alternate Design Method (ACI Code Appendix A); Based on WorkingStress Design method.

13 Places a limit on stresses and uses service loads (ACI A.3).

fcc .45fcfst 20 ksi for grade 40 or 50 steelfst 24 ksi for grade 60 steel

(2.5)

14 Location of neutral axis depends on whether we are analysing or designing a section.

Review: We seek to locate the N.A by taking the first moments:

= Asbd

b(kd) (kd)2 = nAs(d kd)

k =

2n + (n)2 n (2.6)

Design: Objective is to have fc & fs preset & determine As, Fig. 2.4, and we thus seek theoptimal value of k in such a way that concrete and steel reach their respective limitssimultaneously.

cf

fs

d

kd

kd/3C

T

s

c

(1-k/3)d=jd

Figure 2.4: Desired Stress Distribution; WSD Method

cs

= kddkdc =

fcEc

s =fsEs

fcEc

Esfs

= k1kn = EsEcr = fsfc

k = nn+r (2.7)

15 Balanced design in terms of : What is the value of such that steel and concrete will bothreach their maximum allowable stress values simultaneously

C = bkd2 fcT = AsfsC = T

= Asbd

fc2 bkd = bfsbd

k = nn+r

b =

n2r(n+r) (2.8)

16 Governing equations



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Draft2.2 Section Cracked, Stresses Elastic 25Review Start by determining ,

If < b steel reaches max. allowable value before concrete, and

M = Asfsjd (2.9)

If > b concrete reaches max. allowable value before steel and

M = fcbkd

2jd (2.10)

or

M =1

2fcjkbd

2 = Rbd2 (2.11)

where

k = 2n + (n)2 nDesign We define

Rdef=

1

2fckj (2.12)

where k = nn+r , solve for bd2 from

bd2 =M

R(2.13)

assume b and solve for d. Finally we can determine As from

As = bbd (2.14)

17 Summary

Review Design

b,d,As

M

M? b,d,As?

= Asbd k =n

n+r

j = 1 k3k =

2n + (n)2 n r = fsfc

r = fsfc R =12fckj

b =n

2r(n+r)

b =n

2r(n+r) < b M = Asfsjd bd2 = MR

> b M =12fcbkd

2j As = bbd or As =Mfsjd

Example 2-2: Cracked Elastic Section



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Draft26 FLEXURESame problem as example 2.1 fc = 4,000 psi; f

t = 475 psi; fy = 60,000 psi; As = 2.35 in2

however, M is doubled to M = 90 k.ft (instead of 45). Determine concrete and steel stressesSolution:

Based on previous example, fct would be 866 psi fr and the solution is thus no longervalid.

The neutral axis is obtained from

=Asbd

=2.35

(10)(23)= 0.0102 (2.15-a)

n = (0.010)(8) = 0.08174 (2.15-b)

k =

2n + (n)2 n (2.15-c)=

2(0.08174) + (0.08174)2 (0.08174) = 0.33 (2.15-d)

kd = (.33)(23) = 7.6 in (2.15-e)

jd =

1 0.33

3

(23) = 20.47 in (2.15-f)

fs = MAsjd

(2.15-g)

=(90)(1, 000)(12)

(2.35)(20.47)= 22, 400 psi (2.15-h)

fc =2M

bjkd2(2.15-i)

=(2)(90)(12, 000)

(10)(20.47) jd

(7.6)kd

= 1, 390 psi (2.15-j)

I =(10)(7.6)3

12

+ (10)(7.6)7.62 2

+ 8(2.35)(23

7.6)2 = 5, 922 in4 (2.15-k)

Uncracked Cracked Cracked/uncracked

M k.ft 45 90 2N.A in 13.2 7.6fcc psi 485 1,390 (< .5f

c) 2.9I in4 14,710 5,910 .4 ( 1I)fs psi 2,880 22,400 ( 7 ) in 1 4 4

Example 2-3: Working Stress Design Method; Analysis

Same problem as example 2.1 fc = 4,000 psi; f

t = 475 psi; fy = 60,000 psi; As = 2.35 in2.

Determine Moment capacity.Solution:



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Draft2.2 Section Cracked, Stresses Elastic 27

=Asbd

=2.35

(10)(23)= .0102 (2.16-a)

fs = 24 ksi (2.16-b)

fc = (.45)(4, 000) = 1, 800 psi (2.16-c)k =

2n + (n)2 n =

2(.0102)8 + (.0102)2 (8)(.0102) = .331 (2.16-d)

j = 1 k3

= .889 (2.16-e)

N.A. @ (.331)(23) = 7.61 in (2.16-f)

b =n

2r(n + r)=

8

(2)(13.33)(8 + 13.33)= .014 > Steel reaches elastic limit(2.16-g)

M = Asfsjd = (2.35)(24)(.889)(23) = 1, 154 k.in = 96 k.ft (2.16-h)

Note, had we used the alternate equation for moment (wrong) we would have overestimatedthe design moment:

M = =1

2fcbkd

2j (2.17-a)

=1

2(1.8)(10)(0.33)(0.89)(23)2 = 1, 397 k.in > 1, 154 k.in (2.17-b)

If we define c = fc/1, 800 and s = fs/24, 000, then as the load increases both c and sincrease, but at different rates, one of them s reaches 1 before the other.

Load

1 s c

Example 2-4: Working Stress Design Method; Design

Design a beam to carry LL = 1.9 k/ft, DL = 1.0 k/ft with fc = 4, 000 psi, fy = 60, 000 psi,L = 32 ft.Solution:

fc = (.45)(4, 000) = 1, 800 psi (2.18-a)



38/173

Draft28 FLEXUREfs = 24, 000 psi (2.18-b)

n =EsEc

=29, 000

57

4, 000= 8 (2.18-c)

r =fsfc

=24

1.8= 13.33 (2.18-d)

k = nn + r

= 88 + 13.33

= .375 (2.18-e)

j = 1 d3

= 1 .3753

= .875 (2.18-f)

b =n

2r(n + r)=

8

2(13.33)(8 + 13.33)= .01405 (2.18-g)

R =1

2fckj =

1

2(1, 800)(.375)(.875) = 295 psi (2.18-h)

Estimate beam weight at .5 k/ft, thus

M = [(1.9) + (1.0 + .5)] (32)2

8= 435 k.ft (2.19-a)

bd2 =M

R=

435 k.ft in2(12, 000) lb.in

(295) lbs ft k= 17, 700 in3 (2.19-b)

Take b = 18 in & d = 31.4 in h = 36 inCheck beam weight (18)(36)145 (.15)

in2 ft2

in2k

ft3= .675 k/ft

As = (.01405)(18)(31.4) = 7.94 in2 use 8# 9 bars in 2 layers As = 8.00 in2

2.3 Cracked Section, Ultimate Strength Design Method

2.3.1 Whitney Stress Block

b

dh

A

a= cc

C= fabC= fcb c

cf

Actual

c

c

s

1

sf

sf

c

a/2 = c

Figure 2.5: Cracked Section, Limit State

Figure



39/173

Draft2.3 Cracked Section, Ultimate Strength Design Method 2918 At failure we have, linear cross strain distribution (ACI 10.2.2) (except for deep beams),non-linear stress strain curve for the concrete, thus a non-linear stress distribution.

19 Two options:

1. Analytical expression of

exact integration

2. Replace exact stress diagram with a simpler and equivalent one, (ACI 10.2.6)

Second approach adopted by most codes.

20 For the equivalent stress distribution, all we need to know is C & its location, thus and .We adopt a rectangular stress, with depth a = 1c, and stress equal to f

c (ACI 10.2.7.1)

C = fcbc = f

cab (2.20-a)

=favfc

(2.20-b)

a = 1c (2.20-c)

Thus =

1(2.21)

But the location of the resultant forces must be the same, hence

1 = 2 (2.22)

21 From Experiments

fc ( psi)


40/173

Draft210 FLEXURE

C=0.85f abc

h d

b

A

s

s

0.85 fc

c

d

T

=0.003u

a= c1

Figure 2.6: Whitney Stress Block

2.3.2 Balanced Design

Tension Failure:fs = fy

Asfs = .85f

cab = .85f

cb1c

= Asbd

c =

fy.85fc1

d (2.24)

Compression Failure:

c = .003 (2.25-a)

s =fsEs

(2.25-b)

c

d=

.003

.003 + s c = .003fs

Es+.003

d (2.25-c)

Balanced Design:

24 Balanced design occurs if we have simultaneous yielding of the steel and crushing of theconcrete. Hence, we simply equate the previous two equations

fy.85fc1

d = .003fsEs

+.003d

= b

bf2d.85fc1

= .003fsEs

+.003d

Es = 29, 000 ksi

b = .851

fcfy

87,00087,000+fy

(ACI 8.4.3)(2.26)

25 To ensure failure by yielding,

< .75b (2.27)

26 ACI strength requirements

U = 1.4D + 1.7L (ACI 9.2.1)U = 0.75(1.4D + 1.7L + 1.7W) (ACI 9.2.2)

Md = Mu = Mn (ACI 9.1.1) = .90 (ACI 9.3.2.2)

(2.28)



41/173

Draft2.3 Cracked Section, Ultimate Strength Design Method 21127 Also we need to specify a minimum reinforcement ratio

min 200fy

(ACI 10.5.1) (2.29)

to account for temperature & shrinkage

28 Note, that need not be as high as 0.75b. If steel is relatively expensive, or deflection is ofconcern, can use lower .

29 As a rule of thumb, if < 0.5b, there is no need to check for deflection.

2.3.3 Review

30 Given, b,d,As, f

c, fy, determine the moment capacity M.

act =Asbd

b = (.85)1fcfy

87

87+fy

(2.30)

act < b: Failure by yielding and

a =Asfy.85fcb

Fx = 0

Md = Asfy(d a2 ) M = 0(2.31)

act > b is not allowed by code, in this case we have an extra unknown fs.

31 We now have one more unknown fs, and we will need an additional equation (from straindiagram).

c = Asfs.85fcb1Fx = 0

cd =

.003.003+s

From strain diagram

Md = Asfs(d 1c2 ) M = 0(2.32)

We can solve by iteration, or substitution and solution of a quadratic equation.

2.3.4 Design

32 We consider two cases:

I b d and As, unknown; M

dknown; Since design failure is triggered by f

s= f

y

Fx = 0 a =Asfy0.85fcb

= Asbd

a =

fy0.85fc

Md = Asfy

d a2

Md = fy

1 .59 fy

fc

R

bd2(2.33-a)

where is specified by the designer; or

R = fy

1 .59 fy

fc

(2.34)



42/173

Draft212 FLEXUREwhich does not depend on unknown quantities. Then solve for bd2:

bd2 =MdR

(2.35)

Solve for b and d (this will require either an assumption on one of the two, or on theirratio).

As = bd

II b & d known & Md known there is no assurance that we can have a design with bIf the section is too small, then it will require too much steel resulting in an over-reinforcedsection.

Iterative approach

(a) Since we do not know if the steel will be yielding or not, use fs.

(b) Assume an initial value for a (a good start is a = d5 )

(c) Assume initially that fs = fy(d) Check equilibrium of moments (M = 0)

As =Md

fs

d a2 (2.36)

(e) Check equilibrium of forces in the x direction (Fx = 0)

a =Asfs

.85fcb(2.37)

(f) Check assumption offs by either comparing with b, or from the strain diagram

sd c = .003c fs = Es d cc .003 < fy (2.38)

where c = a1 .

(g) Iterate until convergence is reached.

2.4 Practical Design Considerations

2.4.1 Minimum Depth

33 ACI 9.5.2.1 stipulates that the minimum thickness of beams should be

Simply One end Both ends Cantileversupported continuous continuous

Solid Oneway slab L/20 L/24 L/28 L/10

Beams orribbed One way slab L/16 L/18.5 L/21 L/8

where L is in inches, and members are not supporting partitions.

34 Smaller values can be taken if deflections are computed.



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Draft2.4 Practical Design Considerations 2132.4.2 Beam Sizes, Bar Spacing, Concrete Cover

35 Beam sizes should be dimensioned as

1. Use whole inches for overall dimensions, except for slabs use 12 inch increment.

2. Ideally, the overall depth to width ratio should be between 1.5 to 2.0 (most economical).3. For T beams, flange thickness should be about 20% of overall depth.

36 Reinforcing bars

1. Minimum spacing between bars, and minimum covers are needed to

(a) Prevent Honeycombing of concrete (air pockets)

(b) Concrete (usually up to 3/4 in MSA) must pass through the reinforcement

(c) Protect reinforcement against corrosion and fire

2. Use at least 2 bars for flexural reinforcement

3. Use bars #11 or smaller for beams.

4. Use no more than two bar sizes and no more than 2 standard sizes apart (i.e #7 and #9acceptable; #7 and #8 or #7 and #10 not).

5. Use no more than 5 or 6 bars in one layer.

6. Place longest bars in the layer nearest to face of beam.

7. Clear distance between parallel bars not less that db (to avoid splitting cracks) nor 1 in.(to allow concrete to pass through).

8. Clear distance between longitudinal bars in columns not less that 1.5db or 1.5 in.

9. Minimum cover of 1.5 in.

10. Summaries in Fig. 2.7 and Table 2.1, 2.2.

2.4.3 Design Aids

37 Basic equations developed in this section can be easily graphed.

Review Given b d and known steel ratio and material strength, Mn can be readily obtainedfrom Mn = Rbd

2

Design in this case

1. Set Md = Rbd22. From tabulated values, select max and min often 0.5b is a good economical choice.

3. Select R from tabulated values of R in terms of fy, f

c and . Solve for bd2.

4. Select b and d to meet requirements. Usually depth is about 2 to 3 times the width.

5. Using tabulated values select the size and number of bars giving preference to largerbar sizes to reduce placement cost (careful about crack width!).

6. Check from tables that the selected beam width will provide room for the bars chosenwith adequate cover and spacing.



44/173

Draft214 FLEXURE

Bar Nominal Number of Bars

Size Diam. 1 2 3 4 5 6 7 8 9 10#3 0.375 0.11 0.22 0.33 0.44 0.55 0.66 0.77 0.88 0.99 1.10#4 0.500 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00#5 0.625 0.31 0.62 0.93 1.24 1.55 1.86 2.17 2.48 2.79 3.10#6 0.750 0.44 0.88 1.32 1.76 2.20 2.64 3.08 3.52 3.96 4.40#7 0.875 0.60 1.20 1.80 2.40 3.00 3.60 4.20 4.80 5.40 6.00#8 1.000 0.79 1.58 2.37 3.16 3.95 4.74 5.53 6.32 7.11 7.90#9 1.128 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00

#10 1.270 1.27 2.54 3.81 5.08 6.35 7.62 8.89 10.16 11.43 12.70#11 1.410 1.56 3.12 4.68 6.24 7.80 9.36 10.92 12.48 14.04 15.60#14 1.693 2.25 4.50 6.75 9.00 11.25 13.50 15.75 18.00 20.25 22.50

#18 2.257 4.00 8.00 12.00 16.00 20.00 24.00 28.00 32.00 36.00 40.00

Table 2.1: Total areas for various numbers of reinforcing bars (inch2)

Bar Number of bars in single layer of reinf.Size 2 3 4 5 6 7 8

#4 6.8 8.3 9.8 11.3 12.8 14.3 15.8#5 6.9 8.5 10.2 11.8 13.4 15.1 16.7#6 7.0 8.8 10.5 12.3 14.0 15.8 17.5#7 7.2 9.1 11.0 12.8 14.7 16.6 18.5#8 7.3 9.3 11.3 13.3 15.3 17.3 19.3#9 7.6 9.9 12.1 14.4 16.6 18.9 21.2

#10 7.8 10.3 12.9 15.4 18.0 20.5 23.0#11 8.1 10.9 13.7 16.6 19.4 22.2 25.0

#14 8.9 12.3 15.7 19.1 22.5 25.9 29.3#18 10.6 15.1 19.6 24.1 28.6 33.2 37.7

Table 2.2: Minimum Width (inches) according to ACI Code



45/173

Draft2.5 USD Examples 215

Figure 2.7: Bar Spacing

2.5 USD Examples

Example 2-5: Ultimate Strength; Review

Determine the ultimate moment capacity of example 2.1 fc = 4,000 psi; f

t = 475 psi; fy =60,000 psi; As = 2.35 in

2

s

2

t

bA = 2.35 in

25"23"

10"

y

y

Solution:

act =Asbd

=2.35

(10)(23)= .0102 (2.39-a)

b = .851fc

fy

87

87 + fy

= (.85)(.85)4

60

87

87 + 60

= .0285 > act

(2.39-b)

a =Asfy

.85fcb=

(2.35)(60)

(.85)(4)(10)= 4.15 in (2.39-c)

Mn = Asfy

d a2

= (2.35)(60)

23 4.15

2

= 2, 950 k.in (2.39-d)

Md = Mn = 0.9(2, 950) = 2, 660 k.in (2.39-e)

Note:



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Draft216 FLEXURE1. From equilibrium, Fx = 0 c = Asfy.851bfc =

(2.35)(60)(.85)(.85)(4)(10) = 4.87 in

2. Comparing with previous analysis

uncracked cracked ultimate

c 13.2 7.61 4.87M 45 90 2451.7 = 144

3. Alternative solution:

Mn = actfybd2(1 .59act fy

fc) (2.40-a)

= Asfyd(1 59act fyfc

) (2.40-b)

= (2.35)(60)(23)[1 (.59) 604

(.0102)] = 2, 950 k.in = 245 k.ft (2.40-c)

Md = Mn = (.9)(2, 950) = 2, 660 k.in (2.40-d)

Example 2-6: Ultimate Strength; Design I

Design a R/C beam with L = 15 ft; DL = 1.27 k/ft; LL = 2.44 k/ft; fc = 3,000 psi; fy =40 ksi; Neglect beam owns weight; Select = 0.75b

Solution:

wu = 1.4(1.27) + 1.7(2.44) = 5.92 k/ft Factored load (2.41-a)

Md =wuL

2

8=

(5.92)(15)2

8= 166.5 k.ft(12) in/ft = 1, 998 k.in (2.41-b)

= 0.75b = (0.75)(0.85)1fcfy

87

87 + fy(2.41-c)

= (0.75)(.85)23

40

87

87 + 40= .0278 (2.41-d)

R = fy1 .59 fy

fc

(2.41-e)= (.0278)(40)

1 (0.59)(.0278) 40

3

= 0.869 psi (2.41-f)

bd2 =MdR

=1, 998

(0.9)(0.869)= 2, 555 in3 (2.41-g)

Take b = 10 in, d = 16 in As = (.0278)(10)(16) = 4.45 in2 use 3 # 11



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Draft2.5 USD Examples 217

Example 2-7: Ultimate Strength; Design II

Design a R/C beam for b = 11.5 in; d = 20 in; fc = 3 ksi; fy = 40 ksi; Md = 1, 600 k.in

Solution:

Assume a = d5 =205 = 4 in

As =Md

fy(d a2 )=

(1, 600)

(.9)(40)(20 42)= 2.47 in2 (2.42)

check assumption,

a =Asfy

(.85)fcb=

(2.47)(40)

(.85)(3)(11.5)= 3.38 in (2.43)

Thus take a = 3.3 in.

As =(1, 600)

(.9)(40)(20 3.32 )= 2.42 in2 (2.44-a)

a = (2.42)(40)(.85)(3)(11.5)

= 3.3 in

(2.44-b)

act =2.42

(11.5)(20)= .011 (2.44-c)

b = (.85)(.85)3

40

87

87 + 40= .037 (2.44-d)

max = .75b = .0278 > act

(2.44-e)

Example 2-8: Exact Analysis

As an Engineer questioning the validity of the ACI equation for the ultimate flexural capacityof R/C beams, you determined experimentally the following stress strain curve for concrete:

=2

fcmax

1 + max2 (2.45)

where fc corresponds to max.

1. Determine the exact balanced steel ratio for a R/C beam with b = 10, d = 23, fc =4, 000 psi, fy = 60 ksi, max = 0.003.

(a) Determine the equation for the exact stress distribution on the section.

(b) Determine the total compressive force C, and its location, in terms of the locationof the neutral axis c.



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Draft218 FLEXURE(c) Apply equilibrium

2. Using the ACI equations, determine the:

(a) Ultimate moment capacity.

(b) Balanced steel ratio.

3. For the two approaches, compare:

(a) Balanced steel area.

(b) Location of the neutral axis.

(c) Centroid of resultant compressive force.

(d) Ultimate moment capacity.

Solution:

1. Stress-Strain: =

24,000.003

1 +

.003

2 = 2.667 1061 + 1.11 1052 (2.46)2. Assume crushing at failure, hence strain distribution will be given by

=0.003

cy (2.47)

3. Combine those two equation:

=8, 000yc

1 + yc 2 (2.48)

4. The total compressive force is given by

F =

c0

dF = b

c0

dy = b

c0

8, 000 yc

1 +yc

2 dy = b 8, 000cc

0

y

1 +1c

2 dy (2.49)= 8, 000

b

c

1

21c

2 ln

1 +y

c

2c

0

= 8, 000b

c

c2

2ln

1 +

yc

2c0

(2.50)

= 4, 000bc ln(2) = 2, 773bc (2.51)

5. Equilibrium requires that T = C

2, 773bc = Asfy (2.52)

From the strain diagram:

.003

c=

y + .003

d c = (.003)d

y + .003(2.53)

c =(.003)(23)60

29,000 + .003= 13.6 in (2.54)



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Draft2.6 T Beams, (ACI 8.10) 2196. Combining Eq. 2.52 with Eq. 2.54

As =(2, 773)(10)(13.6)

60, 000= 6.28 in2 (2.55)

7. To determine the moment, we must first determine the centroid of the compressive forcemeasured from the neutral axis

y =

ydA

A=

b

ydy

2, 773bc=

b

c0

8, 000 y2

c

1 + (y/c)2

2, 773bcdy =

8, 000b

2, 773bc2

c0

y2

1 +1c

2y2

dy (2.56)

=2.885

c2

c0

y2

1 +1c

2y2

dy =2.885

(13.61)2

y1c

2 11c

2c

0

dy

1 +1c

2y2

(2.57)

= .01557

yc2 c2

1

1c2tan1 y

1

c2

c

0

(2.58)

= .01557

c3 c3 tan1(1) = (.01557)(13.61)3(1 tan1(1)) = 8.43 in (2.59)8. Next we solve for the moment

M = Asfy(d c + y) = (6.28)(60)(23 13.61 8.43) = 6, 713 k.in (2.60)

9. Using the ACI Code

b = .851fcfy

87

87 + 60= (.85)2

4

60

87

147= .0285 (2.61)

As = bbd = (.0285)(10)(23) = 6.55 in

2

(2.62)a =

Asfy.85fcb

=(6.55)(60)

(.85)(4)(10)= 11.57 in (2.63)

M = Asfy

d a2

= (6.55)(60)

23 11.57

Documents

[Victor E Saouma] Machanics and Design of Reinforc(Bookos.org)